pergunta
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| resposta
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Quantas instituições existem?
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CREATE TABLE inst (Id VARCHAR)
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SELECT COUNT(*) FROM inst
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Quantos artigos são publicados no total?
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CREATE TABLE papers (Id VARCHAR)
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SELECT COUNT(*) FROM papers
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Quais são os títulos dos artigos publicados por "Jeremy Gibbons"?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Jeremy" AND t1.lname = "Gibbons"
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Encontre todos os artigos publicados pelo "Aaron Turon".
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Aaron" AND t1.lname = "Turon"
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Quantos artigos publicou o "Atsushi Ohori"?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE papers (paperid VARCHAR)
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SELECT COUNT(*) FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Atsushi" AND t1.lname = "Ohori"
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Qual é o nome da instituição a que pertence "Matthias Blume"?
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR)
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SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Matthias" AND t1.lname = "Blume"
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A que instituição pertence "Katsuhiro Ueno"?
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR)
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SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Katsuhiro" AND t1.lname = "Ueno"
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Quem pertence à instituição "Universidade de Oxford"?
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR)
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SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Oxford"
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Quais autores pertencem à instituição "Google"?
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR)
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SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google"
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Quais são os sobrenomes do autor do artigo intitulado "Binders Unbound"?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR)
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SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Binders Unbound"
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Encontre o nome e o apelido do autor (s) que escreveu o artigo "Não tem nome, não sente dor".
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR)
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SELECT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Nameless , Painless"
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Quais são os trabalhos publicados na instituição "Universidade de Indiana"?
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Indiana University"
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Encontre todos os artigos publicados pela instituição "Google".
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google"
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Quantos trabalhos são publicados pela instituição "Universidade de Tokohu"?
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Tokohu University"
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Encontre o número de artigos publicados pela instituição "Universidade da Pensilvânia".
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Pennsylvania"
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Encontre os papéis que tenham "Olin Shivers" como autor.
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
|
SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Olin" AND t1.lname = "Shivers"
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Que jornais têm "Stephanie Weirich" como autora?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
|
SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Stephanie" AND t1.lname = "Weirich"
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Que artigo é publicado em uma instituição nos "EUA" e tem "Turon" como seu segundo autor?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "USA" AND t2.authorder = 2 AND t1.lname = "Turon"
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Encontre os títulos de artigos cujo primeiro autor é afiliado a uma instituição no país "Japão" e tem o sobrenome "Ohori"?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "Japan" AND t2.authorder = 1 AND t1.lname = "Ohori"
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Qual é o nome do autor que mais trabalhos publicou?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR)
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SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.fname, t1.lname ORDER BY COUNT(*) DESC LIMIT 1
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Retire o país que publicou mais jornais.
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CREATE TABLE inst (country VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR)
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SELECT t1.country FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.country ORDER BY COUNT(*) DESC LIMIT 1
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Encontre o nome da organização que publicou o maior número de artigos.
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CREATE TABLE inst (name VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR)
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SELECT t1.name FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.name ORDER BY COUNT(*) DESC LIMIT 1
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Encontre os títulos dos artigos que contêm a palavra "ML".
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CREATE TABLE papers (title VARCHAR)
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SELECT title FROM papers WHERE title LIKE "%ML%"
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Qual artigo tem o título "Database"?
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CREATE TABLE papers (title VARCHAR)
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SELECT title FROM papers WHERE title LIKE "%Database%"
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Encontre os nomes de todos os autores que escreveram um artigo com o título contendo a palavra "Funcional".
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CREATE TABLE authors (fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR)
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SELECT t1.fname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Functional%"
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Encontre os sobrenomes de todos os autores que escreveram um artigo com título contendo a palavra "monádico".
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR)
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SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Monadic%"
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Retire o título do artigo que tem o maior número de autores.
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CREATE TABLE authorship (authorder INTEGER); CREATE TABLE authorship (paperid VARCHAR, authorder INTEGER); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT MAX(authorder) FROM authorship)
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Qual é o primeiro nome do autor com o sobrenome "Ueno"?
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CREATE TABLE authors (fname VARCHAR, lname VARCHAR)
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SELECT fname FROM authors WHERE lname = "Ueno"
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Encontre o sobrenome do autor com o nome "Amal".
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CREATE TABLE authors (lname VARCHAR, fname VARCHAR)
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SELECT lname FROM authors WHERE fname = "Amal"
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Encontre os nomes dos autores, ordenados por ordem alfabética.
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CREATE TABLE authors (fname VARCHAR)
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SELECT fname FROM authors ORDER BY fname
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Retire todos os sobrenomes dos autores em ordem alfabética.
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CREATE TABLE authors (lname VARCHAR)
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SELECT lname FROM authors ORDER BY lname
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Retirada todos os nomes e apelidos dos autores na ordem alfabética dos apelidos.
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CREATE TABLE authors (fname VARCHAR, lname VARCHAR)
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SELECT fname, lname FROM authors ORDER BY lname
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Quantos nomes diferentes têm os atores e as atrizes?
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CREATE TABLE actor (last_name VARCHAR)
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SELECT COUNT(DISTINCT last_name) FROM actor
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Qual é o primeiro nome mais popular dos atores?
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CREATE TABLE actor (first_name VARCHAR)
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SELECT first_name FROM actor GROUP BY first_name ORDER BY COUNT(*) DESC LIMIT 1
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Qual é o nome completo mais popular dos atores?
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CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR)
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SELECT first_name, last_name FROM actor GROUP BY first_name, last_name ORDER BY COUNT(*) DESC LIMIT 1
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Que distritos têm pelo menos dois endereços?
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CREATE TABLE address (district VARCHAR)
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SELECT district FROM address GROUP BY district HAVING COUNT(*) >= 2
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Qual é o número de telefone e o código postal do endereço 1031 Daugavpils Parkway?
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CREATE TABLE address (phone VARCHAR, postal_code VARCHAR, address VARCHAR)
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SELECT phone, postal_code FROM address WHERE address = '1031 Daugavpils Parkway'
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Qual cidade tem mais endereços?
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CREATE TABLE address (city_id VARCHAR); CREATE TABLE city (city VARCHAR, city_id VARCHAR)
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SELECT T2.city, COUNT(*), T1.city_id FROM address AS T1 JOIN city AS T2 ON T1.city_id = T2.city_id GROUP BY T1.city_id ORDER BY COUNT(*) DESC LIMIT 1
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Quantos endereços há no distrito da Califórnia?
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CREATE TABLE address (district VARCHAR)
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SELECT COUNT(*) FROM address WHERE district = 'California'
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Qual filme é alugado por 0,99 e tem menos de 3 no inventário?
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CREATE TABLE film (title VARCHAR, film_id VARCHAR, rental_rate VARCHAR); CREATE TABLE inventory (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR)
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SELECT title, film_id FROM film WHERE rental_rate = 0.99 INTERSECT SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id HAVING COUNT(*) < 3
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Quantas cidades há na Austrália?
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CREATE TABLE country (country_id VARCHAR, country VARCHAR); CREATE TABLE city (country_id VARCHAR)
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SELECT COUNT(*) FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id WHERE T2.country = 'Australia'
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Que países têm pelo menos 3 cidades?
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CREATE TABLE country (country VARCHAR, country_id VARCHAR); CREATE TABLE city (country_id VARCHAR)
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SELECT T2.country FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id GROUP BY T2.country_id HAVING COUNT(*) >= 3
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Encontre todas as datas de pagamento para os pagamentos com um montante superior a 10 e os pagamentos tratados por um funcionário com o nome de Elsa.
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CREATE TABLE payment (payment_date VARCHAR, staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, first_name VARCHAR); CREATE TABLE payment (payment_date VARCHAR, amount INTEGER)
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SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa'
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Quantos clientes têm um valor ativo de 1?
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CREATE TABLE customer (active VARCHAR)
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SELECT COUNT(*) FROM customer WHERE active = '1'
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Qual é o filme que tem a maior taxa de locação?
|
CREATE TABLE film (title VARCHAR, rental_rate VARCHAR)
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SELECT title, rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1
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Qual filme tem o maior número de atores ou atrizes?
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CREATE TABLE film_actor (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR, description VARCHAR)
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SELECT T2.title, T2.film_id, T2.description FROM film_actor AS T1 JOIN film AS T2 ON T1.film_id = T2.film_id GROUP BY T2.film_id ORDER BY COUNT(*) DESC LIMIT 1
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Qual ator de cinema (atriz) estrelou mais filmes?
|
CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR)
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SELECT T2.first_name, T2.last_name, T2.actor_id FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id ORDER BY COUNT(*) DESC LIMIT 1
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Quais atores de cinema (atriz) desempenharam um papel em mais de 30 filmes?
|
CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR)
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SELECT T2.first_name, T2.last_name FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id HAVING COUNT(*) > 30
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Qual loja possui mais itens?
|
CREATE TABLE inventory (store_id VARCHAR)
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SELECT store_id FROM inventory GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1
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Qual é o montante total de todos os pagamentos?
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CREATE TABLE payment (amount INTEGER)
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SELECT SUM(amount) FROM payment
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Qual cliente, que fez pelo menos um pagamento, gastou menos dinheiro?
|
CREATE TABLE payment (customer_id VARCHAR); CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR)
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SELECT T1.first_name, T1.last_name, T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY SUM(amount) LIMIT 1
|
Qual é o nome do gênero do filme "Hunger Roof"?
|
CREATE TABLE film_category (category_id VARCHAR, film_id VARCHAR); CREATE TABLE film (film_id VARCHAR, title VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR)
|
SELECT T1.name FROM category AS T1 JOIN film_category AS T2 ON T1.category_id = T2.category_id JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'HUNGER ROOF'
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Quantos filmes há em cada categoria?
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CREATE TABLE film_category (category_id VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR)
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SELECT T2.name, T1.category_id, COUNT(*) FROM film_category AS T1 JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T1.category_id
|
Qual é o filme que tem mais cópias no inventário?
|
CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (film_id VARCHAR)
|
SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id ORDER BY COUNT(*) DESC LIMIT 1
|
Qual é o título do filme e o número de inventário do item do inventário que foi alugado com mais frequência?
|
CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (inventory_id VARCHAR, film_id VARCHAR); CREATE TABLE rental (inventory_id VARCHAR)
|
SELECT T1.title, T2.inventory_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T2.inventory_id ORDER BY COUNT(*) DESC LIMIT 1
|
Quantas línguas estão nestes filmes?
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CREATE TABLE film (language_id VARCHAR)
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SELECT COUNT(DISTINCT language_id) FROM film
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- Quais são os filmes classificados como R?
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CREATE TABLE film (title VARCHAR, rating VARCHAR)
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SELECT title FROM film WHERE rating = 'R'
|
Onde fica a loja 1?
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CREATE TABLE store (address_id VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR)
|
SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1
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Qual o pessoal que processou o menor número de pagamentos?
|
CREATE TABLE payment (staff_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR)
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SELECT T1.first_name, T1.last_name, T1.staff_id FROM staff AS T1 JOIN payment AS T2 ON T1.staff_id = T2.staff_id GROUP BY T1.staff_id ORDER BY COUNT(*) LIMIT 1
|
Em que língua é usado o filme AIRPORT POLLOCK?
|
CREATE TABLE film (language_id VARCHAR, title VARCHAR); CREATE TABLE LANGUAGE (name VARCHAR, language_id VARCHAR)
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SELECT T2.name FROM film AS T1 JOIN LANGUAGE AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'AIRPORT POLLOCK'
|
Quantas lojas há?
|
CREATE TABLE store (Id VARCHAR)
|
SELECT COUNT(*) FROM store
|
Quantos tipos de classificações diferentes estão listados?
|
CREATE TABLE film (rating VARCHAR)
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SELECT COUNT(DISTINCT rating) FROM film
|
Que filmes têm 'Escenas apagadas' como substring no recurso especial?
|
CREATE TABLE film (title VARCHAR, special_features VARCHAR)
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SELECT title FROM film WHERE special_features LIKE '%Deleted Scenes%'
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Quantos itens no inventário tem a loja 1?
|
CREATE TABLE inventory (store_id VARCHAR)
|
SELECT COUNT(*) FROM inventory WHERE store_id = 1
|
Quando foi o primeiro pagamento?
|
CREATE TABLE payment (payment_date VARCHAR)
|
SELECT payment_date FROM payment ORDER BY payment_date LIMIT 1
|
Onde vive a cliente com o nome de Linda e qual é o e-mail dela?
|
CREATE TABLE customer (email VARCHAR, address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR)
|
SELECT T2.address, T1.email FROM customer AS T1 JOIN address AS T2 ON T2.address_id = T1.address_id WHERE T1.first_name = 'LINDA'
|
Encontre todos os filmes com mais de 100 minutos, ou classificados PG, exceto aqueles que custam mais de 200 para a substituição.
|
CREATE TABLE film (title VARCHAR, replacement_cost INTEGER, LENGTH VARCHAR, rating VARCHAR)
|
SELECT title FROM film WHERE LENGTH > 100 OR rating = 'PG' EXCEPT SELECT title FROM film WHERE replacement_cost > 200
|
Qual é o nome e o apelido do cliente que fez o primeiro aluguel?
|
CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR); CREATE TABLE rental (customer_id VARCHAR, rental_date VARCHAR)
|
SELECT T1.first_name, T1.last_name FROM customer AS T1 JOIN rental AS T2 ON T1.customer_id = T2.customer_id ORDER BY T2.rental_date LIMIT 1
|
Qual é o nome completo do funcionário que alugou um filme a um cliente com o nome de April e o sobrenome Burns?
|
CREATE TABLE customer (customer_id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE rental (staff_id VARCHAR, customer_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR)
|
SELECT DISTINCT T1.first_name, T1.last_name FROM staff AS T1 JOIN rental AS T2 ON T1.staff_id = T2.staff_id JOIN customer AS T3 ON T2.customer_id = T3.customer_id WHERE T3.first_name = 'APRIL' AND T3.last_name = 'BURNS'
|
Qual loja tem mais clientes?
|
CREATE TABLE customer (store_id VARCHAR)
|
SELECT store_id FROM customer GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1
|
Qual é o montante de pagamento mais elevado?
|
CREATE TABLE payment (amount VARCHAR)
|
SELECT amount FROM payment ORDER BY amount DESC LIMIT 1
|
Onde vive a funcionária com o nome de Elsa?
|
CREATE TABLE staff (address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR)
|
SELECT T2.address FROM staff AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T1.first_name = 'Elsa'
|
Quais são os primeiros nomes dos clientes que não alugaram nenhum filme após "2005-08-23 02:06:01"?
|
CREATE TABLE customer (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER); CREATE TABLE rental (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER)
|
SELECT first_name FROM customer WHERE NOT customer_id IN (SELECT customer_id FROM rental WHERE rental_date > '2005-08-23 02:06:01')
|
Quantas filiais bancárias há?
|
CREATE TABLE bank (Id VARCHAR)
|
SELECT COUNT(*) FROM bank
|
Quantos clientes há?
|
CREATE TABLE bank (no_of_customers INTEGER)
|
SELECT SUM(no_of_customers) FROM bank
|
Encontre o número de clientes nos bancos de Nova Iorque.
|
CREATE TABLE bank (no_of_customers INTEGER, city VARCHAR)
|
SELECT SUM(no_of_customers) FROM bank WHERE city = 'New York City'
|
Encontre o número médio de clientes em todos os bancos do estado de Utah.
|
CREATE TABLE bank (no_of_customers INTEGER, state VARCHAR)
|
SELECT AVG(no_of_customers) FROM bank WHERE state = 'Utah'
|
Encontre o número médio de clientes em todos os bancos.
|
CREATE TABLE bank (no_of_customers INTEGER)
|
SELECT AVG(no_of_customers) FROM bank
|
Encontre a cidade e o estado da filial do banco chamada Morningside.
|
CREATE TABLE bank (city VARCHAR, state VARCHAR, bname VARCHAR)
|
SELECT city, state FROM bank WHERE bname = 'morningside'
|
Encontre os nomes das agências bancárias no estado de Nova York.
|
CREATE TABLE bank (bname VARCHAR, state VARCHAR)
|
SELECT bname FROM bank WHERE state = 'New York'
|
Escrever o nome de todos os clientes, ordenados por saldo da conta, em ordem ascendente.
|
CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR)
|
SELECT cust_name FROM customer ORDER BY acc_bal
|
Escrever o nome de todos os clientes que têm algum empréstimo, ordenado por seu valor total do empréstimo.
|
CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR)
|
SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount)
|
Encontre o estado, tipo de conta e pontuação de crédito do cliente cujo número de empréstimo é 0.
|
CREATE TABLE customer (state VARCHAR, acc_type VARCHAR, credit_score VARCHAR, no_of_loans VARCHAR)
|
SELECT state, acc_type, credit_score FROM customer WHERE no_of_loans = 0
|
Encontre o número de cidades em que os bancos estão localizados.
|
CREATE TABLE bank (city VARCHAR)
|
SELECT COUNT(DISTINCT city) FROM bank
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Encontre o número de estados em que os bancos estão localizados.
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CREATE TABLE bank (state VARCHAR)
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SELECT COUNT(DISTINCT state) FROM bank
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Quantos tipos diferentes de contas há?
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CREATE TABLE customer (acc_type VARCHAR)
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SELECT COUNT(DISTINCT acc_type) FROM customer
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Encontre o nome e o saldo da conta do cliente cujo nome inclui a letra a.
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CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR)
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SELECT cust_name, acc_bal FROM customer WHERE cust_name LIKE '%a%'
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Encontre o saldo total de cada cliente de Utah ou Texas.
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CREATE TABLE customer (acc_bal INTEGER, state VARCHAR)
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SELECT SUM(acc_bal) FROM customer WHERE state = 'Utah' OR state = 'Texas'
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Encontre o nome dos clientes que têm os dois tipos de contas de poupança e de cheques.
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CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR)
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SELECT cust_name FROM customer WHERE acc_type = 'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type = 'checking'
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Encontre o nome dos clientes que não têm uma conta poupança.
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CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR)
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SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type = 'saving'
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Encontre o nome de clientes que não têm um empréstimo com um tipo de hipotecas.
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CREATE TABLE loan (cust_id VARCHAR, loan_type VARCHAR); CREATE TABLE customer (cust_name VARCHAR); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR)
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SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Mortgages'
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Encontre o nome de clientes que têm empréstimos de ambos os empréstimos hipotecários e de automóveis.
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CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR)
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SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Auto'
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Encontre o nome dos clientes cuja pontuação de crédito é inferior à média de todas as pontuações de crédito dos clientes.
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CREATE TABLE customer (cust_name VARCHAR, credit_score INTEGER)
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SELECT cust_name FROM customer WHERE credit_score < (SELECT AVG(credit_score) FROM customer)
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Encontre o nome da filial do banco que tem o maior número de clientes.
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CREATE TABLE bank (bname VARCHAR, no_of_customers VARCHAR)
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SELECT bname FROM bank ORDER BY no_of_customers DESC LIMIT 1
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Encontre o nome do cliente que tem a pontuação de crédito mais baixa.
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CREATE TABLE customer (cust_name VARCHAR, credit_score VARCHAR)
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SELECT cust_name FROM customer ORDER BY credit_score LIMIT 1
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Encontre o nome, o tipo de conta e o saldo da conta do cliente que tem a pontuação de crédito mais alta.
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CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR, acc_bal VARCHAR, credit_score VARCHAR)
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SELECT cust_name, acc_type, acc_bal FROM customer ORDER BY credit_score DESC LIMIT 1
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Encontre o nome do cliente que tem o maior montante de empréstimos.
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CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR)
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SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount) DESC LIMIT 1
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Encontre o estado que tem mais clientes.
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CREATE TABLE bank (state VARCHAR, no_of_customers INTEGER)
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SELECT state FROM bank GROUP BY state ORDER BY SUM(no_of_customers) DESC LIMIT 1
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Para cada tipo de conta, encontre o saldo médio da conta dos clientes com pontuação de crédito inferior a 50.
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CREATE TABLE customer (acc_type VARCHAR, acc_bal INTEGER, credit_score INTEGER)
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SELECT AVG(acc_bal), acc_type FROM customer WHERE credit_score < 50 GROUP BY acc_type
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Para cada estado, encontrar o saldo total das contas dos clientes cuja pontuação de crédito seja superior a 100.
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CREATE TABLE customer (state VARCHAR, acc_bal INTEGER, credit_score INTEGER)
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SELECT SUM(acc_bal), state FROM customer WHERE credit_score > 100 GROUP BY state
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